Catenaries, Parabolas and Suspension Bridges

CATENARIES, PARABOLAS AND SUSPENSION BRIDGES 

DAVID GRIFFIN 

SUSPENSION BRIDGES / D A GRIFFIN

TALK CONTENTS 

A Brief, Pictorial History of Suspension Bridges. 

‣ Famous Examples 

‣ Problems and Solutions. 

Galileo 

‣ The Equations for the Chain Curve of a Suspension Bridge. 

‣ The Equation for a Hanging Chain. Parabola or Catenary? 

The Catenary and Jakob Bernoulli’s Challenge. 

‣ Leibniz 

‣ Hugens 

‣ Bernoulli 

The Methods for Deriving the Equation for the Catenary. 

‣Calculus. 

‣Differential Equation. 

‣The Calculus of Variations. 

The Relationship Between the Parabola and the Catenary. 

The Inverted Parabola. 

‣Arch Bridges 

The Inverted Catenary. 

‣Arches 

‣Can Catenaries Help You to Cycle with Square Wheels? 


A “Suspended-deck Suspension Bridge” 


The remains of the Maya Bridge at Yaxchilan, Mexican/Guatemalan border. 

The earliest known suspension-deck suspension bridge. 100m in three spans. 7 th Century. 


Faust Vrančić 

In 1595 the Croation bishop Faust Vrančić designed a 

suspension bridge, but it was never constructed. 


James Creek Suspension Bridge, Pennsylvania. 

James Finlay, 1801. 

Bridge demolished 1833. 

The first modern suspension bridge. 

It used wrought-iron cables. 


Union Bridge, River Tweed, 1820. 

The oldest suspension bridge still 

carrying traffic. 

Dryburgh Bridge, River Tweed. 

Opened1817. Collapsed 1818. 


The Menai Suspension Bridge 

Thomas Telford, Completed 1826 


The Clifton Suspension Bridge 

I K Brunel, Completed1864. 



Two cables 

support one 

deck. 


BROOKLYN BRIDGE 

John Augustus Roebling,1883. 


Four cables, two decks. 


A light walkway is suspended between the two decks of the Brooklyn Bridge. 


The Golden Gate Suspension Bridge 

Irving Morrow, Charle Alton Ellis, Leon Moissieff, 1937. 


PC 

LT 

Tacoma Narrows Bridge 

Leon Moissieff. Opened1940. 


Map 

New Tacoma Narrows Bridge (Background: Mt Rainier) 

Charles E Andrew and Dexter R Smith,1950 


A second bridge at Tacoma Narrows was built in 2007. 


The Severn Bridge 

William Brown, 1966 


Brown designed the Severn Bridge to avoid 

the problems of the Tacoma Bridge. It has 

a slender, aerodynamic deck. 


The Humber Bridge 

John Hyatt, Douglas Strachan and others, 1981. 


THE TOP TEN SUSPENSION BRIDGES 

Suspension bridge are typically ranked by the length of their main span. 

Akashi-Kaikyo Bridge (Japan) 1991 m 1998 

Xihoumen Bridge (China) 1650 m 2007 

Great Belt Bridge (Denmark) 1624 m 1998 

Runyang Bridge (China) 1490 m 2005 

Humber Bridge (England) 1410 m 1981 

(The longest span from 1981 until 1998.) 

Jiangyin Suspension Bridge (China) 1385 m 1997 

Tsing Ma Bridge (Hong Kong), 1377 m 1997 

(Longest span with both road and metro.) 

Verrazano-Narrows Bridge (USA) 1298 m 1964 


Golden Gate Bridge (USA) 1280 m 1937 


Yangluo Bridge (China) 1280 m 2007 


Millenium Bridge 2000 

Arup, Foster and Partners 

PC 

LT 

Synchronous Lateral Excitation 

http://www2.eng.cam.ac.uk/~den/ICSV9_06.htm 


Dampers on Millenium Bridge to prevent synchronous lateral excitation. 


The first person to study the physics and mathematics of the 

suspension bridge was Galileo. 

Galileo Galilei 

1564 - 1642 



SUSPENSION BRIDGE: FORCES 

Cable 

y 

Tension at P. 

T 

P 

θ 

Tension at O. 

T o 

0 

Section of cable between O and P. 

P has horizontal coordinate x. 

Deck 

Weight of red 

section of deck. 

W 

Section of deck supported by 

cables segment OP. Length = x. 


y 

y 

T 

T 

T 

T o 

0 

P 

θ 

P 

θ 

x 

T o 

W 

x 

W 

The three forces T 0 

, T and W are in equilibrium. 

T 

W 

They form a triangle of forces with tanθ = W / T o 

. 

θ 

T is tangential to the chain at P. 

T o 


y 

y 

T 

T 

T 

W 

T o 

0 

P 

θ 

P 

θ 

x 

θ 

T o 

W 

x 

δy 

θ 

The triangle of forces is similar to the differential 

triangle at P. 

They both have gradient tanθ = W / T o 

. 

δx 


δy 

T 

W 

θ 

θ 

Gradient 

y 

 

x 

δx 

Gradient 

W 

T O 

T o 

y 

x 

W 

T O 

x = horizontal distance from the point O. 

W xg 

mass / length 

y 

xg 

 

x 

T O 

g 

k 

T O 

y kx 

x 

In the limit: 

dy 

dx 

kx 

μ has the dimensions of mass/length: m.l -1 

g has the dimensions of acceleration: l.t -2 

T o 

has the dimensions of force: m.l.t -2 

Thus k has the dimensions of length -1 : l -1 


THE EQUATION FOR THE CHAIN-CURVE OF A SUSPENSION BRIDGE 

dy 

 

dx 

kx 

2 

kx 

y kx. dx 

y C 2 

Since y = 0 when x = 0 C must = 0 

2 

kx 

y 

2 

A Parabola 

k has the dimensions of length -1 . 

x has the dimensions of length. 

Thus: 

y has the dimensions of length. 


Tension in the Cable 

T 

μxg 

θ 

T o 

T 

 

2 2 2 

g 

x 

T O 

The tension in the cable is greatest at the towers. 

The tension in the cable is a minimum at the lowest point and = T o 

. 


Other Applications of Suspension-bridge Technology 

BRASSIERES: 

An Engineering Miracle 

From Science and Mechanics, February, 1964 

By Edward Nanas 

“There is more to brassiere design than meets the eye. In many respects, the challenge of 

enclosing and supporting a semi-solid mass of variable volume and shape, plus its adjacent 

mirror image - together they equal the female bosom - involves a design effort comparable to that 

of building a bridge or a cantilevered skyscraper. “ 

http://www.firstpr.com.au/show-and-tell/corsetry-1/nanas/engineer.html 


THE CATENARY 

Catenary, Alysoid, Chainette. 



A spider’s web: multiple catenaries. 


“Simple suspension bridges” 

or rope bridges are catenaries, 

not parabolas. 

Söderskär Bridge, Finland. 


GALILEO AND THE CATENARY 

Galileo believed that a catenary had the equation of a parabola. 

He had studied the parabola in various contexts and was the first 

to state that a projectile would follow the path of a parabola. 

In 1669 a posthumous publication by Joachim Jungius proved that 

the function describing a catenary could not be algebraic and could 

not therefore be a parabola. 

Joachim Jungius 

1587 – 1657. 


THE JAKOB BERNOULLI CHALLENGE. 

In 1690 Jakob Bernoulli issued a challenge to Leibniz, Christiaan Huygens 

and Johann Bernoulli to derived the equation for the catenary. 

The solutions were presented in 1691. 

Gottfried Leibniz 

Christiaan Huygens 

Johann Bernoulli 

Newton also solved the 

problem: anonymously. 

The Age of Big Hair. 

Jakob Bernoulli 


The Bernoulli Family Tree 

Several generations of 

mathematical geniuses. 

Jakob 

Johann (I) 

Daniel 

Nicolaus (II) 

Johann (II) 

Johann (III) 

Nicolaus (III) 


Jakob Bernoulli 

(1654-1705) 

First studied to be a minister. 

Studied at Basel University. 

‣ Received degree in theology. 

‣ Fascinated by mathematics. 

Furthered the calculus he had 

learned from Leibniz. 

Studied catenaries. 

Worked on the design of bridges. 

Studied the brachistochrone 

problem with Johann. 

Was a professor at Basel until 

his death. 


THE SOLUTIONS TO THE BERNOULLI CHALLENGE. 

Leibniz used calculus, but did not show his method. 

Johann Bernoulli used the calculus of variations. 

‣This involves finding the shape which minimizes 

the potential energy of the system. 

Huygens used a complicated geometric proof. 

A solution using differential equations can also used. 

In 1691, when the derivation of the equation for the catenary was 

published, the Jesuit priest Ignace Gaston Paradies published a 

text-book on forces and geometry which included the derivation of 

the equations for the suspension-bridge cable and the catenary. 




Cable 

y 

T 

Tension at P. 

P 

θ 

Tsinθ 

Tension at O. 

T o 

0 

s 

Tcosθ 

W = μsg 


Cable 

y 

T 

Tension at P. 

T 

P 

θ 

Tsinθ 

Tension at O. 

T o 

0 

s 

Tcosθ 

T o 

W = μsg 

µsg 

The three forces T 0 

, T and µsg are in equilibrium. 

They form a triangle of forces with tanθ = µsg / T o 

. 

Since T is tangential to the curve formed by the chain 

tanθ is equal to the gradient at the point P. 

θ 

T 

T o 

µsg 

Gradient = µsg / T o 

. 


Derivation of the Catenary Equation Method 1 

We have shown: 

tan 

sg 

T O 

s 

 

TO 

g 

tan 

Define: 

k 

 

T 

O 

g 

T o 

has the dimension of force: m.l.t -2 

μ has the dimension of mass/length: m.l -1 

g has the dimension of acceleration: l.t -2 

Thus: 

k has the units of length: l 

s k tan 



Consider the differential triangle. 

In the limit δs approaches the value 

of the hypotenuse. 

δS 

θ 

δy 

dx 

ds 

cos 

δx 

The above triangle is similar to the 

triangle of forces. 

T 

µsg 

θ 

T o 



s k tan 

δS 

δy 

ds 

2 

k sec 

d 

(1) 

θ 

δx 

From (1) and (2): 

dx 

 

d 

dx 

ds 

 

ds 

d 

dx 

ds 

cos 

(2) 

dx 

 

d 

k.sec 

2 

.cos 

 

k sec 

dx 

d 

k sec 



s k tan 

δS 

δy 

ds 

2 

k sec 

d 

(1) 

θ 

δx 


dy dy ds 

 

d ds d 

dy 

ds 

sin 

(2) 

dy 

d 

k sec 2 .sin 

dy 

d 

k sec.tan 



Deriv 

dx 

d 

k sec 

 

 

x 

dx 

ksec . 

d 

Separate variables and integrate 

y 

x k.ln sec 

tan 

C 

T 

x 

 

0, 

 

0 

θ 

0 k.ln1 

0 C C 0 

P 

0 

x 

k. ln sec tan 

x 

Parametric equation of the catenary (1) 



dy 

d 

k sec.tan 

 

 

y dy k sec.tan. 

d 

Separate variables and integrate 

y 

 

k sec 

C 

We have not defined where the axis y = 0 is. 

Define C = 0. 

Thus when θ = 0 y = k. 

} 

k 

x 

y 

k sec 

Parametric equation of the catenary (2) 



We now have the two parametric equations for the catenary. 

We need to eliminate θ to obtain the cartesian x-y equation. 

x 

k. ln sec tan 

y k sec 

(1) (2) 

y 

From (2): sec (3) 

k 

2 

1 

tan sec 

2 

 

tan 

 

sec 

2 

1 

(4) 

From (1), (3) and (4): 

x 

 

k.ln 

y 

k 

 

 

 

 

y 

k 

 

 

 

2 

1 

1 

2 

Recall the identity: cosh u ln u u 1 



2 

y y 

1 

2 

x k.ln 

1 

cosh u ln u u 1 

(1) 

k k 

(2) 


x 

 

k cosh 

1 

 

 

 

y 

k 

 

 

 

Rearranging and taking the 

cosh function of both sides 

of the equation gives: 

 

cosh 

 

x 

 

k 

 

y 

k 

y 

 

 

k cosh 

 

x 

 

k 

If x has the dimensions of length: 

‣ x/k is dimensionless. 

‣ y has the dimensions of length. 

Jungius was correct. The catenary is not described 

by an algebraic function; and is thus not a parabola. 


The Derivation of the Catenary by Differential Equations. 

We have previously shown: 

δS 

θ 

δx 

δy 


s k tan 

(1) 

dy 

tan (2) 

dx 

dy 

s k. 

dx 

2 

ds d y k. dx 

2 

dx 



δS 

θ 

δx 

δy 

ds 2 dx 2 

dy 2 

ds 

dx 

 

dy 

1 

 

dx 

2 

(1) 

We have just shown: 

ds 

dx 

k 

2 

d y 

. dx 

2 

(2) 


2 

d y dy 

k. 1 

2 

 

dx dx 

2 




2 

d y dy 

k. 1 

2 

 

dx dx 

2 

(1) 

Let: 

y' 

 

dy 

dx 

(2) 


 

 

dy' 

k. 1 

y 

dx 

' 2 

Separate variables 

2 dx 

d y' 

 

1 

y' 

. 

k 

dx 

k 

 

d 

1 

y' 

 

y' 

2 

 

dx 

k 

 

 

d 

1 

y' 

 

y' 

 

2 


dx 

k 

 

 

d 

1 

y' 

 

y' 

 

2 

(1) 

Recall the standard integral: 

du 1 

1 

u 

2 

sinh 

u 

(2) 

 

From (1) and (2): y' 

C 

x 

k 

sinh 1 When x 0 y' 0 

 

Thus: C 0 

Take sinh function of both sides: 

x 

k 

 

 

sinh 1 y' 

 

x 

sinh 

y' 

k 

 

dy 

dx 

x 

sinh 

 

k 

Separate the variables 

y 

 

 

k cosh 

 

x 

k 

 

 

 

C 

x 

dy . dx 

k 

 

sinh 

 

 

Once again we can define the 

coordinate axes so that C = 0. 

y 

 

 

k cosh 

 

x 

 

k 


The Relationship Between 

the Parabola and the Catenary 


Online function plotter: http://www.mathe-fa.de/en#anchor 

A Comparison of a Parabola and a Catenary 

SUSPENSION BRIDGES / D A GRIFFIN 

y 

x 

2.cosh 

2 

2 

 

y 

x 

2 

2

The Relationship Between the Parabola and the Catenary. 

The MacLaurin Series for a Catenary 

k cosh 

x 

k 

cosh 

x 

k 

 

0 

k 

2n 

2n 

x 

.(2n)! 

2 4 6 

cosh x x x x 

1 

........ 

2 4 6 

k k .2! k .4! k .6! k 

2 

2 4 6 

cosh x x x x 

k k(1 

........ 

2 4 6 

k k .2! k .4! k .6! k 

2 

2n 

x 

........ 

.(2n)! 

n 

2n 

x 

........) 

.(2n)! 

2 4 6 

2n 

cosh x x x x x 

k k ........ ........) 

3 5 

2n 

k k.2! 

k .4! k .6! k 

1 .(2n)! 

n 

k cosh 

x 

k 

2 

x 

2 k 

k 

If k is >1 the catenary can be approximated 

by a parabolic function for small values of x. 


If k is >1 the catenary can be approximated by a parabolic function for small values of x. 

(k = 2) 

x 

y 2.cosh 

2 

y 

2 

x 

4 

2 


The Relationship Between the Parabola and the Catenary 

If the parabola y = x 2 is rolled along the x-axis the locus 

of its focus is the catenary: 

1 

y cosh 4x 

4 


The Construction of the Brooklyn Bridge 

Parabola and Catenary 

During the construction of the Brooklyn Bridge 

it was possible at one stage to contrast a laden 

and an un-laden cable: a parabola and a catenary. 








Inverted Parabolas and Catenaries 

Arch Bridges 

Free-standing Arches 


The New River Gorge Bridge, Virginia. 

A supported-deck bridge. 


The Tyne Bridge 

A compression-arch suspended-deck bridge 


SUPPORTED ARCH BRIDGE 

The forces acting on a section of the arch are compressive. 

An analysis of the triangle of forces leads once again to a parabola. 

The arch is rigid. I does not assume the shape of an inverted parabola. 

It should be constructed as an inverted parabola if it is to have a uniform 

deck supported at regular intervals. 

The triangle of forces acting on a segment is analogous to that for a suspension bridge. 

y 

W 

C O 

x 

C O 

θ 

C 

W = μxg 

C 


THE INVERTED CATENARY 

The inverted catenary is the ideal curve for 

an arch which supports only its own weight. 

It is the minimum energy structure. 

The forces are primarily of compression. 


. 

St Louis Gateway Arch, 

Eero Saarinen, Completed 1965. 


St Louis Gateway Arch 


This formula is inscribed on the arch. 

Thus when x = 0 and y is at a maximum 

y = 630 ft. 

This also gives a separation of 630 ft 

for the bases. 



Taq–i-Kisra, Ctesiphon, Mesopotamia / Irak 


Casa Milà, Barcelona. 

Antoni Gaudí 



Antoni Gaudí 



Antoni Gaudí 


Can you ride 

a cycle with 

square wheels? 

http://www.maa.org/mathland/mathtrek_04_05_04.html 


For the rolling square the shape of the road is a series 

of inverted, truncated catenary curves. 

PC 

LT 


For regular n-sided polygonal wheels the curve of the road 

is made from inverted catenaries with the equation: 

y = - Rcot(/n).cosh(x/A) 


In practice a triangular wheel would get stuck. 

The vertices puncture the road. 

Road 

“Spoke” 

Bottom of triangular wheel 


www.exploratorium.edu/texnet/exhibits/motion/square.../square_cbk.pdf 


Proving that the locus of the centre of a square as it 

rolls over an inverted catenaries is a straight line. 

There are several, diverse proofs. Some are long and complicated. 

www.maa.org/pubs/mathmag.html 

(General study on “roads” and non-circular “wheels”.) 

www.macalester.edu/mathcs/documents/catenaries.pdf 

(Uses two coordinate systems: polar and cartesian.) 

http://www.maplesoft.com/applications/view.aspx?SID=6322 

(Method using several differential equations.) 

http://www.snc.edu/math/squarewheelbike.html 

Follow hypertext link on the website mathematics. 

(Uses standard geometry and calculus: but long.) 

See: Wikipedia page on Roulette (curve). 

(Generates the catenary road as a roulette in the complex plane.) 


Consider the square resting 

on a vertex and symmetrically 

poised between two of the 

humps. 

We want the centre of the 

circle to remain on the line 

y = a√2 as the square rolls 

along the road. 

y 

2a 

Y = a√2 

a√2 

Road, y = f(x) 

x 


y 

2a 

C 

θ 

y = a√2 

B 

θ 


A 

x 


a.sec y a 

2 

y 

dy 

dx 

 

tan 

2a 

C 

θ 

y = a√2 

a.secθ 

a 

B 

θ 


Appendix 

y 

A 

SUSPENSION BRIDGES / D A GRIFFIN 

x

a.sec y a 

2 

dy 

dx 

tan 

2 2 

1tan sec 

 

sec 2 

2 

y 

a 

(1) 

dy 

2 

1 

sec 

 

dx 

 

(2) 

Let: 

u sec 

(3) 


u 

2 

y 

a 


dy 

 

1 

 

dx 

 

2 

u 

2 

(4) 

dy 

dx 

u 

2 

1 


dy 2 

u 1 

dx u 2 

y 

a 

du du dy 

 

dx dy dx 

du 

dy 

1 

 

a 

du 

dx 

1 

a 

2 

 

u 

1 

Separate variables. 

1 

dx 

a 

u 

du 

2 

1 


1 

dx 

a 

u 

du 

2 

1 

 

1 

dx 

a 

 

u 

du 

2 

1 

x 

a 

x 

u C 

 

a 

1 

C 

cosh 

u 

cosh 

u 

2 

y 

a 

Take cosh function of both sides. 

x 

y a 2 a.cosh 

 

C 

 

a 

y x 

2 coshC 

 

a a 

When x = 0 y = f(x) = 0. Therefore: 

0 

0a 2 a.coshC 

 

a 

 

2 cosh C 

 

C cosh 2 

1 

 

 

ya a 

 

a 

1 

2 cosh cosh 2 x 

 

The equation of the road. 


Plots of the Inverted Catenary “Road” and Related Functions. 

1 

x 

y a 2 acoshcosh 2 

 

a 

Road equation 

y cosh( x) 

y 

y 

 

1 

cosh cosh 2 

 

x 

1 

2 cosh cosh 2 

(Road equationtion with a = 1) 

 

x 

 


1 

y a 2 acoshcosh 2 

 

x 

 

a 

Solve for y = 0 and dy/dx = 0 

y 

(Values when a = 1) 

 

1 

a.cosh 2, a 2 a 

 

(0.8814, 0.4142) 

 

1 

2.cosh a 2,0 

 

(1.7627, 0) 

(0,0) 

x 


PC 

PC 

LT 

LT 


APPENDIX 1 

Proof that the Arc Length of the Inverted Catenaries is 

Equal to the Length of a Side of the Square: i.e. 2a. 

dx 

 

s 

1 . dy 

dy 

 

2 

dy 

tan 

dx 

dx 

cot 

dy 

(1) (2) (3) 

 

2 

From (1) and (3): s 1cot . dy 

cos . 

s ec dy 

(4) 

a.sec y a 

2 

y a 2 asec 

dy 

a sec .tan 

d 

(5) 


s a.cos ec .sec .tan . 

d 

2 

s a.sec . 

d 

 

 


y 

2a 

a√2 

45 o 45 o 135 o 

x 


2 

s a.sec . 

d 

 

 

2 

2 

s a.sec . 

d 

 

 

1 

 

2 

s a tan 

C 

1 

1 45 o 

2 135 o 

tan 45 1 

tan135 1 

s a1 1 2a 

 

 


APPENDIX 2 

TNB 


APPENDIX 3 

Conversion of FLV (e.g. Youtube) files to downloadable files (e.g. MP4). 


APPENDIX 4 

http://www.mathe-fa.de/en 


APPENDIX 5 

1 2 

cosh x ln x x 1 

e 

cosh x 

e 

sinh x 

x 

x 

e 

2 

e 

2 

x 

x 

cosh x sinh x e 

x 

2 2 

cosh x sinh x 1 

y 

cosh 

1 

x 

x 

cosh 

y 

cosh 

y 

sinh y e 

y 

2 2 

cosh y 

sinh y 1 

y 

2 

e cosh y cosh y1 

e x x 

y 

2 

 

1 

y x x 

2 

ln 1 

1 2 

cosh x ln x x 1

Catenaries, Parabolas and Suspension Bridges

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